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Math Help - Integration: Trigonometric substitution

  1. #1
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    Integration: Trigonometric substitution

    Hey

    The problem i am stuck on is: find the integral of (x^2)/(sqrt(9-x^2))

    I have managed to complete the problem to the second to last step where i get (9 theta)/2-9/4 sin(2 theta),

    since i substituted at the beginning x=3sin(theta) i can rearrange for theta and get the first part = 9/2sin^-1(x/3).
    What I am struggling with is to complete the 9/4sin(2theta) by putting this back in terms of x.
    The answer i am looking for is 1/2 x sqrt(9-x^2) so if someone could explain how to get 9/4 sin(2 theta) to 1/2 x sqrt(9-x^2) that be be extremely helpful.

    Thanks
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Sam1111 View Post
    Hey

    The problem i am stuck on is: find the integral of (x^2)/(sqrt(9-x^2))

    I have managed to complete the problem to the second to last step where i get (9 theta)/2-9/4 sin(2 theta),

    since i substituted at the beginning x=3sin(theta) i can rearrange for theta and get the first part = 9/2sin^-1(x/3).
    What I am struggling with is to complete the 9/4sin(2theta) by putting this back in terms of x.
    The answer i am looking for is 1/2 x sqrt(9-x^2) so if someone could explain how to get 9/4 sin(2 theta) to 1/2 x sqrt(9-x^2) that be be extremely helpful.

    Thanks
    \tfrac{9}{4}\sin\!\left(2\theta\right)=\tfrac{9}{4  }\left(2\sin\theta\cos\theta\right)=\tfrac{9}{2}\s  in\theta\cos\theta. From your substitution, we know that x=3\sin\theta\implies \sin\theta=\frac{x}{3}. With this, it is easily shown that \cos\theta=\frac{\sqrt{9-x^2}}{3}.

    Thus, \tfrac{9}{2}\sin\theta\cos\theta=\tfrac{9}{2}\left  (\tfrac{1}{3}x\cdot\tfrac{1}{3}\sqrt{9-x^2}\right)=\tfrac{1}{2}x\sqrt{9-x^2}

    Does this make sense?
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